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Solubility Product

In a saturate...

Question

In a saturated solution of the sparingly soluble strong electrolyte $A g I O_{3}$ (Molecular mass = 283) the equilibrium which sets in is
$A g I O_{3 (s)} ⇌ A g_{(a q)}^{+} + I O_{3}^{-} (a q)$ .
If the solubility product constant $K_{s p}$ of $A g I O_{3}$ at a given temperature is $1.0 \times 10^{- 8},$ what is the mass of $A g I O_{3}$ contained in 100 mL of its saturated solution?

A

28.3 \times 10^{- 2} g

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B

2.83 \times 10^{- 3} g

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C

1.0 \times 10^{- 7} g

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D

1.0 \times 10^{- 4} g

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Solution

The correct option is B $2.83 \times 10^{- 3} g$
Let the solubility of $A g I O_{3}$ be 's'
The dissociation equation is
$A g I O_{3} (s) ⇌ A g^{+} (a q) + I O_{3}^{-} (a q) t = 0100 a t e q b 1 - s s s$
$K_{s p} = [A g^{+}] [I O_{3}^{-}] K_{s p} = s^{2}$
$s^{2} = 1.0 \times 10^{- 8}$
$s = 10^{- 4} m o l L^{- 1}$
Mass of $A g I O_{3}$ in 100 mL
$= \frac{c o n c e n t r a t i o n \times M o l a r m a s s}{1000} \times V o l u m e = \frac{10^{- 4} \times 283}{1000} \times 100 = 0.00283$
Mass of $A g I O_{3} = 2.83 \times 10^{- 3} g$

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Solubility Product

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